# How can I think of the flat space metric tensor as a multilinear function?

I'm pretty new to the idea of tensors, and I'm having a bit of confusion with how to think about the flat space metric tensor in special relativity.

I understand that a good way to think about tensors is as multilinear functions, which takes a number of vectors from a vector space $$V$$ over a field $$F$$ as arguments, and returns a scalar from $$F$$.

However, I don't really know how to think about the flat space / Minkowski metric tensor in this way. In my class, I understand it's come up in the equation

$$ds^2 = \eta_{\alpha \beta} dx^\alpha dx^\beta$$

where $$\eta_{\alpha \beta}$$ is the metric tensor, and $$dx^\alpha$$ is the $$\alpha$$'th component of the differential 4-vector. I just don't really see how we can think of $$\eta$$ as a multilinear function here. Can anyone shed any light on this?

Thanks!

• The equation you wrote is equivalent to $ds^2 = \eta(dx, dx)$. So you see that the metric is a function that takes in two arguments and returns a number, and furthermore from the equation you wrote, it is linear in each argument. – knzhou Sep 28 at 20:18

This longer answer is a supplement to tparker's good concise answer.

The notation $$dx^a$$ is used for a couple of different-but-related things:

• First perspective: Intuitively, we often think of it as an infinitesimal displacement. In this view, $$ds^2=g_{ab}dx^a dx^b$$ is giving us the infinitesimal change $$ds$$ in proper time or proper distance (depending on the sign convention) along an infinitesimal segment of a worldline. In this perspective, the equation $$ds^2=g_{ab}dx^a dx^b$$ is really an abbreviation for $$\left(\frac{ds}{d\lambda}\right)^2 =g_{ab}\frac{dx^a}{d\lambda} \frac{dx^b}{d\lambda}$$ where $$\lambda$$ is a parameter that runs along a given worldline so that the $$\lambda$$th point has coordinates $$x^a(\lambda)$$.

• Second perspective: More mathematically, $$dx^a$$ is a one-form, which takes a vector field as input and returns its $$a$$th component as output. In terms of coordinates, a vector field $$V$$ is a linear combination of partial derivatives, $$V=V^a\partial_a$$, and the one-form $$dx^a$$ has the property $$dx^a(V) = V^a$$. Equivalently, $$dx^a(\partial_b)=\delta^a_b$$.

The second perspective most directly explains why people say that the metric is a bilinear function. The metric $$g=g_{ab}dx^a dx^b$$ takes two vectors fields as input and returns a scalar field as output, like this (see tparker's answer): $$g(A,B) = g_{ab}dx^a(A) dx^b(B) = g_{ab} A^a B^b.$$ To relate this to the first perspective, think of the given worldline as one member of a congruence of non-overlapping worldlines that covers the whole spacetime (within a given region), all parameterized by $$\lambda$$ in such a way that $$\lambda$$ can be viewed as a new coordinate in some new coordinate system. The vector field $$\partial/\partial\lambda$$ is everywhere tangent to those worldlines. In terms of the original coordinate system, this vector field is $$\frac{\partial}{\partial\lambda} = \frac{dx^a}{d\lambda} \frac{\partial}{\partial x^a},$$ so its components are $$dx^a\left(\frac{\partial}{\partial\lambda}\right) = \frac{dx^a}{d\lambda}.$$ Use this to get $$g\left(\frac{\partial}{\partial\lambda}, \,\frac{\partial}{\partial\lambda}\right) =g_{ab}\frac{dx^a}{d\lambda} \frac{dx^b}{d\lambda},$$ which shows the relationship between the two perspectives. Note the two different usages of the notation $$dx^a$$, sometimes as a one-form and sometimes as the numerator of a derivative (that is, as an infinitesimal displacement).

This all applies in flat spacetime just as it does in curved spacetime, because this all applies whether or not the components $$g_{ab}$$ are themselves functions of the coordinates.

The metric $$\eta$$ is a bilinear function that inputs two four-vectors (call them $$A$$ and $$B$$) and outputs the scalar $$A \cdot B = A^\mu B_\mu = A^\mu \eta_{\mu \nu} B^\nu$$. It is clearly bilinear (i.e. separately linear in both the first argument $$A$$ and the second argument $$B$$) by the distributive property.